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$\newcommand{\tensor}{\otimes}$ $\newcommand{\lieg}{\mathfrak{g}}$ $\newcommand{\iso}{\cong}$

Tensor product of two unitary modules

The tensor product of two unitary modules $V_1$ and $V_2$ over an associative commutative ring $A$ with a unit is the $A$-module $V_1 \tensor_A V_2$ together with an $A$-bilinear mapping

$$(x_1, x_2) \mapsto x_1 \tensor x_2 \in V_1 \tensor_A V_2$$ which is universal in the following sense: For any $A$-bilinear mapping $\beta: V_1 \times V_2 \to W$, where $W$ is an arbitrary $A$-module, there is a unique $A$-linear mapping $b : V_1 \tensor_A V_2 \to W$ such that

$$\beta(x_1, x_2) = b(x_1 \tensor x_2), \qquad x_1 \in V_1, \qquad x_2 \in V_2.$$ The tensor product is uniquely defined up to a natural isomorphism. It always exists and can be constructed as the quotient module of the free $A$-module $F$ generated by the set $V_1 \times V_2$ modulo the $A$-submodule $R$ generated by the elements of the form

$$(x_1 + y, x_2) - (x_1, x_2) - (y, x_2),$$

$$(x_1, x_2 + z) - (x_1, x_2) - (x_1, z),$$

$$(cx_1, x_2) - c(x_1, x_2),$$

$$(x_1, cx_2) - c(x_1, x_2),$$

$$x_1, y \in V_1, \qquad x_2, z \in V_2, \qquad c \in A;$$ then $x_1 \tensor x_2 = (x_1, x_2) + R$. If one gives up the requirement of commutativity of $A$, a construction close to the one described above allows one to form from a right $A$-module $V_1$ and a left $A$-module $V_2$ an Abelian group $V_1 \tensor_A V_2$, also called the tensor product of these modules [1]. In what follows $A$ will be assumed to be commutative.

The tensor product has the following properties:

$$A \tensor_A V \iso V,$$

$$V_1 \tensor_A V_2 \iso V_2 \tensor_A V_1,$$

$$(V_1 \tensor_A V_2) \tensor V_3 \iso V_1 \tensor_A (V_2 \tensor_A V_3),$$

$$\left( \bigoplus_{i \in I} V_i \right) \tensor_A W \iso \bigoplus_{i \in I} (V_i \tensor_A W)$$ for any $A$-modules $V$, $V_i$ and $W$.

If $(x_i)_{i \in I}$ and $(y_j)_{j \in J}$ are bases of the free $A$-modules $V_1$ and $V_2$, then $(x_i \tensor y_j)_{(i,j) \in I\times J}$ is a basis of the module $V_1 \tensor_A V_2$. In particular,

$$\dim(V_1 \tensor_A V_2) = \dim V_1 \cdot \dim V_2$$ if the $V_i$ are free finitely-generated modules (for instance, finite-dimensional vector spaces over a field $A$). The tensor product of cyclic $A$-modules is computed by the formula

$$(A/I) \tensor_A (A/J) \iso A/(I+J)$$ where $I$ and $J$ are ideals in $A$.

One also defines the tensor product of arbitrary (not necessarily finite) families of $A$-modules. The tensor product

$$\bigotimes^p V = V \tensor_A \cdots \tensor_A V \qquad (p \text{ factors})$$ is called the $p$-th tensor power of the $A$-module $V$; its elements are the contravariant tensors (cf. Tensor on a vector space) of degree $p$ on $V$.

To any pair of homomorphisms of $A$-modules $\alpha_i : V_i \to W_i$, $i=1,2$, corresponds their tensor product $\alpha_1 \tensor \alpha_2$, which is a homomorphism of $A$-modules $V_1 \tensor_A V_2 \to W_1 \tensor_A W_2$ and is defined by the formula

$$(\alpha_1 \tensor \alpha_2) (x_1 \tensor x_2) = \alpha(x_1)\tensor \alpha_2(x_2), \qquad x_i \in V_i.$$ This operation can also be extended to arbitrary families of homomorphisms and has functorial properties (see Module). It defines a homomorphism of $A$-modules $$\Hom_A(V_1, W_1) \tensor_A \Hom_A(V_2, W_2) \to \Hom_A(V_1 \tensor V_2, W_1 \tensor W_2),$$ which is an isomorphism if all the $V_i$ and $W_i$ are free and finitely generated.


Comments

An important interpretation of the tensor product in (theoretical) physics is as follows. Often the states of an object, say, a particle, are defined as the vector space $V$ over $\C$ of all complex linear combinations of a set of pure states $e_i$, $i \in I$. Let the pure states of a second similar object be $f_j$, $j \in J$, yielding a second vector space $W$. Then the pure states of the ordered pair of objects are all pairs $(e_i, f_j)$ and the space of states of this ordered pair is the tensor product $V\tensor_\C W$.

Tensor product of two algebras

The tensor product of two algebras $C_1$ and $C_2$ over an associative commutative ring $A$ with a unit is the algebra $C_1 \tensor_A C_2$ over $A$ which is obtained by introducing on the tensor product $C_1 \tensor_A C_2$ of $A$-modules a multiplication according to the formula

$$(x_1 \tensor x_2)(y_1 \tensor y_2) = (x_1 y_1) \tensor (x_2 y_2), \qquad x_i, y_i \in C_i.$$ This definition can be extended to the case of an arbitrary family of factors. The tensor product $C_1 \tensor_A C_2$ is associative and commutative and contains a unit if both algebras $C_i$ have a unit. If $C_1$ and $C_2$ are algebras with a unit over the field $A$, then $\widetilde C_1 = C_1 \tensor \mathbf{1}$ and $\widetilde C_2 = \mathbf{1} \tensor C_2$ are subalgebras of $C_1 \tensor_A C_2$ which are isomorphic to $C_1$ and $C_2$ and commute elementwise. Conversely, let $C$ be an algebra with a unit over the field $A$, and let $C_1$ and $C_2$ be subalgebras of it containing its unit and such that $x_1 x_2 = x_2 x_1$ for any $x_i \in C_i$. Then there is a homomorphism of $A$-algebras $\phi : C_1 \tensor_A C_2 \to C$ such that $\phi(x_1 \tensor x_2) = x_1 x_2$, $x_i \in C_i$. For $\phi$ to be an isomorphism it is necessary and sufficient that there is in $C_1$ a basis over $A$ which is also a basis of the right $C_2$-module $C$.

Tensor product of two matrices (by D.A. Suprunenko)

The tensor product, or Kronecker product (cf. Matrix multiplication), of two matrices $A = [ \alpha_{ij} ]$ and $B$ is the matrix

$$A \tensor B = \begin{bmatrix} \alpha_{11} B & \cdots & \alpha_{1n} B \\ \vdots & \ddots & \vdots \\ \alpha_{m1} B & \cdots & \alpha_{mn} B \end{bmatrix}.$$ Here, $A$ is an $(m\times n)$-matrix, $B$ is a $(p \times q)$-matrix and $A \tensor B$ is an $(mp \times nq)$-matrix over an associative commutative ring $k$ with a unit.

Properties of the tensor product of matrices are:

$$(A_1 + A_2) \tensor B = A_1 \tensor B + A_2 \tensor B,$$

$$A \tensor (B_1 + B_2) = A \tensor B_1 + A\tensor B_2,$$

$$\alpha(A \tensor B) = \alpha A \tensor B = A \tensor \alpha B,$$ where $\alpha \in k$,

$$(A \tensor B)(C \tensor D) = AC \tensor BD.$$ If $m=n$ and $p=q$, then

$$\det(A \tensor B) = (\det A)^p (\det B)^n.$$ Let $k$ be a field, $m=n$ and $p=q$. Then $A\tensor B$ is similar to $B \tensor A$, and $\det(A \tensor E_p - E_n \tensor B)$, where $E_s$ is the unit matrix, coincides with the resultant of the characteristic polynomials of $A$ and $B$.

If $\alpha : V \to V'$ and $\beta : W \to W'$ are homomorphisms of unitary free finitely-generated $k$-modules and $A$ and $B$ are their matrices in certain bases, then $A \tensor B$ is the matrix of the homomorphism $\alpha \tensor \beta : V \tensor W \to V' \tensor W'$ in the basis consisting of the tensor products of the basis vectors.

Tensor product of two representations (by A.I. Shtern)

The tensor product of two representations $\pi_1$ and $\pi_2$ of a group $G$ in vector spaces $E_1$ and $E_2$, respectively, is the representation $\pi_1 \tensor \pi_2$ of $G$ in $E_1 \tensor E_2$ uniquely defined by the condition

$$(\pi_1 \tensor \pi_2) (g) (\xi_1 \tensor \xi_2) = \pi_1(g) \xi_1 \tensor \pi_2(g) \xi_2 \tag{*}$$ for all $\xi_1 \in E_1$, $\xi_2 \in E_2$ and $g \in G$. If $\pi_1$ and $\pi_2$ are continuous unitary representations of a topological group $G$ in Hilbert spaces $E_1$ and $E_2$, respectively, then the operators $(\pi_1 \tensor \pi_2)(g)$, $g \in G$, in the vector space $E_1 \tensor E_2$ admit a unique extension by continuity to continuous linear operators $(\pi_1 \tensor -\pi_2)g$, $g\in G$, in the Hilbert space $E_1 \tensor -E_2$ (being the completion of the space $E_1 \tensor E_2$ with respect to the scalar product defined by the formula $(\xi_1 \tensor \xi_2, \eta_1 \tensor \eta_2) = (\xi_1, \eta_1)(\xi_2, \eta_2)$) and the mapping $\pi_1 \tensor \pi_2 : g \to (\pi_1 \tensor -\pi_2)g$, $g \in G$, is a continuous unitary representation of the group $G$ in the Hilbert space $E_1 \tensor -E_2$, called the tensor product of the unitary representations $\pi_1$ and $\pi_2$. The representations $\pi_1 \tensor \pi_2$ and $\pi_2 \tensor \pi_1$ are equivalent (unitarily equivalent if $\pi_1$ and $\pi_2$ are unitary). The operation of tensor multiplication can be defined also for continuous representations of a topological group in topological vector spaces of a general form.


Comments

If $\pi_i$ is a representation of an algebra $A_i$ in a vector space $E_i$, $i=1,2$, one defines the tensor product $\pi_1 \tensor \pi_2$, which is a representation of $A_1\tensor A_2$ in $E_1\tensor E_2$, by

$$(\pi_1 \tensor \pi) (a_1 \tensor a_2) = \pi_1(a_1) \tensor \pi_2(a_2).$$ In case $A = A_1 = A_2$ is a bi-algebra (cf. Hopf algebra), composition of this representation with the comultiplication $A \to A \tensor A$ (which is an algebra homomorphism) yields a new representation of $A$, (also) called the tensor product.

In case $G$ is a group, a representation of $G$ is the same as a representation of the group algebra $k[G]$ of $G$, which is a bi-algebra, so that the previous construction applies, giving the same definition as (*) above. (The comultiplication on $k[G]$ is given by $g\mapsto g \tensor g$.)

In case $\lieg$ is a Lie algebra, a representation of $\lieg$ is the same as a representation of its universal enveloping algebra, $U_\lieg$, which is also a bi-algebra (with comultiplication defined by $x\mapsto 1 \tensor x + x \tensor 1$, $x \in \lieg$). This permits one to define the tensor product of two representations of a Lie algebra:

$$(\pi_1 \tensor \pi_2)(x) = 1 \tensor \pi_2(x) + \pi_1(x) \tensor 1.$$

Tensor product of two vector bundles

The tensor product of two vector bundles $E$ and $F$ over a topological space $X$ is the vector bundle $E\tensor F$ over $X$ whose fibre at a point $x \in X$ is the tensor product of the fibres $E_x \tensor F_x$. The tensor product can be defined as the bundle whose transfer function is the tensor product of the transfer functions of the bundles $E$ and $F$ in the same trivializing covering (see Tensor product of matrices, above).


Comments

For a vector bundle $E$ over a space $X$ and a vector bundle $F$ over a space $Y$ one defines the vector bundle $E \times F$ over $X \times Y$ (sometimes written $E \tensor F$) as the vector bundle over $X \times Y$ with fibre $E_x \tensor F_y$ over $(x, y)$. Pulling back this bundle by the diagonal mapping $x \mapsto (x, x)$ defines the tensor product defined above.


References

[1] N. Bourbaki, "Elements of mathematics. Algebra: Algebraic structures. Linear algebra" , 1 , Addison-Wesley (1974) pp. Chapt.1;2 (Translated from French)
[2] F. Kasch, "Modules and rings" , Acad. Press (1982) (Translated from German)
[3] A.I. Kostrikin, Yu.I. Manin, "Linear algebra and geometry" , Gordon & Breach (1989) (Translated from Russian)
[4] P.R. Halmos, "Finite-dimensional vector spaces" , v. Nostrand (1958)
[5] M.F. Atiyah, "$K$-theory: lectures" , Benjamin (1967)
How to Cite This Entry:
Tensor product. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Tensor_product&oldid=43382
This article was adapted from an original article by A.L. Onishchik (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article